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Let $R$ be a ring (not commutative in general) with identity and let $\Psi(x) = x^m-\sum_{j=0}^{m-1}\psi_jx^j$ be a monic polynomial over $R$. I want to construct a ring extension $K$ of $R$, which contains a root of $\Psi(x)$. One construction follows from non-commutative Hamilton-Caley Theorem.

But I also think that there is another construction, which seems to be very interesting.

Let $M = R[x]/R[x]\Psi(x)$ be a left module over $R[x]$ and $A = \mathrm{Ann}_{R[x]}M\subset R[x]$ be an annihilator of $M$. I want to prove that $\Psi(x)$ has a root in the ring $$ R[x]/A. $$

The proof is very simple in the case, where $R$ is a commutative ring. So the main interest is when $R$ is not commutative.

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  • $\begingroup$ So you want to show that $\Psi(x + A) = \Psi(x) + A$. Isn't $x$ central in $R[x]$? $\endgroup$
    – Luc Guyot
    Commented Aug 20, 2017 at 11:42
  • $\begingroup$ Yes, x is central in R[x]. $\endgroup$
    – Mikhail Goltvanitsa
    Commented Aug 20, 2017 at 11:45
  • $\begingroup$ We certainly have $\Psi(x + A) + A = \Psi(x) + A$, but $\Psi(x) \in A$ holds if and only if $\Psi(x) R[x] \subset R[x] \Psi(x)$. So I would expect a negative answer. $\endgroup$
    – Luc Guyot
    Commented Aug 20, 2017 at 12:58
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    $\begingroup$ If $a$ and $b$ are two non-commuting elements of $R$ and $\Psi(x) = x - a$ then $\Psi(x)(x - b) \in \Psi(x) R[x] \setminus R[x] \Psi(x)$. So $\Psi(x) \notin A$. Therefore $x + A$ is not a root of $\Psi$ in $R[x]/A$. But you are asking about any root in $R[x]/A$? $\endgroup$
    – Luc Guyot
    Commented Aug 20, 2017 at 13:21
  • $\begingroup$ @LucGuyot, yes, x+A is not a root of $\Psi(x)$ in general case (but in commutative is). Moreover, my experiments show that the root can be choosen in the form $x+H(x)\Psi(x)+A$ for some polynomyal $H(x)$. $\endgroup$
    – Mikhail Goltvanitsa
    Commented Aug 20, 2017 at 17:17

1 Answer 1

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No, $R[x]/A$ does not always have a root.

Let $R=\mathbb{Z}\langle a,b\rangle$ be the free ring on two noncommuting elements $a$ and $b$. Let $\Psi=x^2+a\in R[x]$, and let $M$ and $A$ be the objects you defined.

In this case, $A=0$. Suppose $f=r_nx^n+\cdots+r_0\in A$. In particular, there is some $g=s_n x^n+\cdots+s_0\in R[x]$ such that $f\cdot b=g\cdot (x^2+a)$. Matching up coefficients, we find that $r_0b=s_0a$, $r_1b=s_1a$, $r_2b=s_2a+s_0$,$\dots$, $r_ib=s_ia+s_{i-2}$. In the free ring $R$ the first two equations force $r_0=s_0=0$ and $r_1=s_1=0$, and by induction the rest of the equations force $f=g=0$. Hence $A=0$.

Finally, the ring $R[x]/A\simeq R[x]$ doesn't have any roots of $\Psi$.

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